H_
Matrix Representations
SKIP_
p1
QM_[0;S;not defined;not defined;T;W;A;not defined;defined but not listed;not defined]
Recall that the product of an m by n matrix A and an n by p matrix B
is defined to be the m by p matrix whose entry in row i and column j
is the dot product of row i of matrix A and column j of matrix B. Note
that this is defined only if A has the same number of columns as B
has rows and that the operation is not commutative. Evaluate the products
where
and
AH_[2]
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
AB
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
BA
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
AC
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
CA
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
AD
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
AI
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
IA
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
AZ
AS_[A;B;C;D;I;Z;R;S;T;U;V;W;not defined;defined but not listed]
ZA
SKIP_
p2
QM_[0;6;12;3;6]
Let g be the linear map from
to itself defined by f(1) = 6.
AH_[2]
Assuming the standard bases for the domain and codomain, the matrix
for g is (
AC_[4]
).
Assuming the standard basis for the codomain and the basis (2) for
the domain, the matrix for g is (
AC_[4]
).
Assuming the standard basis for the domain and the basis (2) for
the codomain, the matrix for g is (
AC_[4]
).
Assuming the bases for the domain and the codomain are (2),
the matrix for g is (
AC_[4]
).
SKIP_
p3
QM_[0;3;1;-5;4;2;-1;-9;13;1;6;-2;5;-5;11;-7;12]
Let f be the linear map from
to
defined by
f((1,0)) = (3,-5) and f((0, 1)) = (1, 4). (As usual, I will often write
column vectors as row vectors -- you are expected to interpret them
correctly from the context.)
AH_[2]
Assuming that one is using the standard bases, the matrix of f is
AC_[4]
AC_[4]
AC_[4]
AC_[4]
Assume that the basis for the codomain is the standard basis but that the
basis for the domain is (1, -1), (-1, 2) (in this order). The matrix of
f is then
AC_[4]
AC_[4]
AC_[4]
AC_[4]
Assume that the basis for the domain is the standard basis but that the
basis for the codomain is (1, -1), (-1, 2) (in this order). The matrix of
f is then
AC_[4]
AC_[4]
AC_[4]
AC_[4]
Assume that the bases for the domain and the codomain are
(1, -1), (-1, 2) (in this order). The matrix of
f is then
AC_[4]
AC_[4]
AC_[4]
AC_[4]
SKIP_
p4
QM_[0;2;-9;-1;13;2;-1;-9;13;2;-1;-9;13;2;-1;-9;13]
Suppose the linear map f from
to
is defined
by the matrix
where the
matrix was defined with respect to the basis
(in this order)
for the codomain and
(in this order) for the domain.
Suppose the linear map g from
to
is defined
by the matrix
where the
matrix was defined with respect to the basis
(in this order)
for the codomain and
(in this order) for the domain. Let
h = f(g) be the composition of the two linear maps.
AH_[2]
One has
AC_[4]
AC_[4]
and
AC_[4]
AC_[4]
.
The matrix of the map h with respect to the basis
for the codomain
and
for the domain is:
AC_[4]
AC_[4]
AC_[4]
AC_[4]
The product AB is
AC_[4]
AC_[4]
AC_[4]
AC_[4]
The matrix of the map f with respect to the basis
(in this order) is
AC_[4]
AC_[4]
AC_[4]
AC_[4]
SKIP_
p5
QM_[0;m;n;n;p;true;false;true;true;true]
Let
and
be linear maps of finite
dimensional vector spaces. Let
,
, and
be bases of U, V, and W respectively. Let
and
be
the matrices of f and g respectively, where each matrix is written using
the above bases (in the order specified). Let h = f(g).
AH_[2]
The matrix A has
AS_[m;n;p;none of these]
rows and
AS_[m;n;p;none of these]
columns. The matrix B has
AS_[m;n;p;none of these]
rows and
AS_[m;n;p;none of these]
columns.
AS_[true;false]
For each index i = 1, ..., p, one has
.
AS_[true;false]
For each index i = 1, ..., n, one has
.
AS_[true;false]
For each index i = 1, ..., p, one has
.
AS_[true;false]
The entry in row k and column i of the matrix AB is
.
AS_true;false]
The matrix of the composition h = f(g) written with respect to the
above bases of W and U is the matrix AB.
SKIP_
p6
QM_[0;true;false]
Answer the following questions either true or false; make sure you
can prove the assertion or give a counter-example:
AH_[2]
AS_[true;false]
Because composition of functions is associative, i.e. f(g(h)) = (f(g))(h),
we know that multiplication of matrices is associative.
If A is a matrix corresponding to a one-to-one linear map and there
is a matrix B with BA equal to an identity matrix, then A is non-singular
with inverse B.
AS_[true;false]
SKIP_
p7
QM_[0;true;true;true;false;true;true]
Let A be any m by n matrix with real entries. Recall that applying
an elementary row operation to A is the same as multiplying on the
left by a particular non-singular matrix. Let I be the m by m identity
matrix (which has 1 along the main diagonal and 0 elsewhere).
AH_[2]
AS_[true;false]
Swapping rows i and j of A amounts to multiplying it on the left by the
matrix obtained by swapping rows i and j and the matrix I.
AS_[true;false]
Multiplying row i of A by s amounts to multiplying A on the the left by
the matrix obtained by multiplying row i of I by s.
AS_[true;false]
Replacing row i of A with the sum of row i and m times row j
amounts to multiplying A on the left by the matrix obtained from I by
replacing row i of A with the sum of row i and m times row j of I.
AS_[true;false]
Suppose that applying row operation
to a matrix is equivalent to
multiplying it on the left by the matrix
for i = 1, 2, ..., k.
Then the result of applying the operations
(in this
order) to the matrix is equivalent to multiplying the matrix on the left
by the product
.
AS_[true;false]
Suppose that applying row operation
to a matrix is equivalent to
multiplying it on the left by the matrix
for i = 1, 2, ..., k.
Then the result of applying the operations
(in this
order) to the matrix is equivalent to multiplying the matrix on the left
by the product
.
Now suppose that A is a square m by m matrix. Form the matrix
B = (A | I), i.e. B has m rows and 2n columns, the first m columns of
B are the same as the corresponding columns of A, and the remaining
m columns are those of the m by m identity matrix. Use Gaussian
elimination to reduce B to its reduced echelon form C. Then C = DB
where D is obtained as in the last part. It is
AS_[true;false]
that if A is non-singular, the C = (I | D) for some matrix D. It follows
from C = DB and B = (A | I) that DA = I and so D is the inverse of A.
SKIP_
p8
QM_[0.000001;4/17;-1/17;5/17;3/17]
Use the method suggested by the last problem to find the inverse
of the matrix
.
(Enter your answer as exact fractions, e.g. 4/3 rather than a decimal
approximation.)
AH_[2]
AC_[4]
AC_[4]
AC_[4]
AC_[4]
SKIP_
p9
QM_[0.00001;5/17;-6/17;2/17;1/17;5/17;-6/17;2/17;1/17]
Let f be the linear map of
to itself defined by the
matrix
where we are
using the standard basis. Consider the basis
.
AH_[2]
The matrix of f using the standard basis for the domain and the new
basis for the codomain is exactly
AC_[4]
AC_[4]
AC_[4]
AC_[4]
Let
, then using the results of the
last exercise, one can calculate
to be exactly
AC_[4]
AC_[4]
AC_[4]
AC_[4]
SKIP_
p10
QM_[0;0;4;1;4;2]
Consider the matrices corresponding to the following linear maps from
the vector space of polynomials of degree at most n to itself. The
bases are always the standard one:
.
AH_[2]
For the map f(p(x)) = p(2x), the entry in the 3rd row and 4th column
is
AC_[4]
.
For the map f(p(x)) = p(2x), the entry in the 3rd row and 3th column
is
AC_[4]
.
For the map f(p(x)) = p(x + 2), the entry in the 3rd row and 3th column
is
AC_[4]
.
For the map f(p(x)) = p(x + 2), the entry in the 2nd row and 3th column
is
AC_[4]
.
For the map f(p(x)) = p'(x), the entry in the 2nd row and 3th column
is
AC_[4]
.